Question

The solution of the differential equation $(x+2y^3)\frac{dy}{dx}=y$ is

• A. $y^3+Cx=y$
• B. $\frac{x{y}^4}{2}+xy=Cy$
• C. $y^3+Cy=x$
• D. $x+2y^3=y+C$

Answer 1

Answer: (C)

$y^3+Cy=x$

Given, $(x+2y^3)\frac{dy}{dx}=y$
$\Rightarrow y\frac{dx}{dy}=x+2y^3$
$\Rightarrow \frac{dx}{dy}-\frac{1}{y}x=2y^2$
This is of the form $\frac{dx}{dy}+Px=Q$
where, $P=-\frac{1}{y}$ and $Q=2y^2$
Thus, the given equation is linear.
$\therefore IF=e^{\int Pdy}=e^{\int -\frac{1}{y}dy}=e^{-\log y}=e^{\log {\left(y\right)}^{-1}}=y^{-1}=\frac{1}{y}$
So, the required solution is
$x\cdot IF=\int (Q\cdot IF)dy+C$
$\Rightarrow x\cdot \frac{1}{y}=\int (2y^2\cdot \frac{1}{y})dy+C$
$\Rightarrow x\cdot \frac{1}{y}=\int 2ydy+C=y^2+C$
$\Rightarrow x=y^3+Cy$.